Question: Evaluate $\int^{\pi/4}_0\tan^4x\,dx\,$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac13+\dfrac\pi4$ (Choice B) B $\dfrac43+\dfrac\pi4$ (Choice C) C $-\dfrac23+\dfrac\pi4$ (Choice D) D $-\dfrac23$
Answer: The integrand has even powers of $~\tan x\,$. We will use the identity $~\tan^2x=\sec^2x-1~$ to get appropriate powers of $~\tan x~$ and its derivative $~\sec^2x~$ together. $ \begin{aligned}\int^{\pi/4}_0\tan^4x\,dx&=\int^{\pi/4}_0\tan^2x\cdot\tan^2x\,dx\\ \\ \\&=\int^{\pi/4}_0(\sec^2x-1)\tan^2x\,dx\\ \\ \\&=\int^{\pi/4}_0\tan^2x\sec^2x\,dx-\int^{\pi/4}_0\tan^2x\,dx\\ \\ \\&=\int^{\pi/4}_0\tan^2x\sec^2x\,dx-\int^{\pi/4}_0\big(\sec^2x-1\big)\,dx\end{aligned}$ For the first integral we use a $~u$ -substitution. Let $ u=\tan x~~~~~$ and $~~~~~du=\sec^2x\,dx\,$. For this integral, we also need to change the limits of integration. We leave the second integral in unchanged in terms of $~x\,$. $ x=0~~~~~\Rightarrow~~~~~u=0$ $ x=\dfrac\pi4~~~~~\Rightarrow~~~~~u=1$ $ \begin{aligned}\int^{\pi/4}_0\tan^4x\,dx&=\int^{\pi/4}_0\tan^2x\sec^2x\,dx-\int^{\pi/4}_0\big(\sec^2x-1\big)\,dx\\ \\ \\&=\int_0^1 u^2\,du-\int_0^{\pi/4}\big(\sec^2x-1\big)\,dx\end{aligned}$ Now we can evaluate the definite integral. $ \begin{aligned}\int^{\pi/4}_0\tan^4x\,dx&=\int_0^1 u^2\,du-\int_0^{\pi/4}\big(\sec^2x-1\big)\,dx\\ \\ \\&=\frac{u^3}3\Bigg|_0^1~~-\Big(\tan x-x\Big)\Bigg|_0^{\pi/4}\\ \\ \\&=\Big(\frac13-0\Big)-\Big(\big(1-\frac\pi4\big)-\big(0-0\big)\Big)\\ \\ \\&=\dfrac13-1+\dfrac\pi4\\ \\ \\&= -\dfrac23+\dfrac\pi4\end{aligned}$